Linear Motion Under a Variable Force (9231 Further Mechanics 3.5)
Hello! Welcome to one of the most exciting (and perhaps challenging) topics in Further Mechanics. In AS/A Level Mechanics (9709), you focused mostly on motion where the acceleration was constant—like gravity or simple friction. We used the familiar SUVAT equations.
In this chapter, we step up! We deal with forces that change dynamically—perhaps they depend on how fast the object is moving, or where the object is located. Because the force changes, the acceleration changes, and we must use Calculus (specifically, differential equations) to solve the motion.
Don't worry if this sounds tricky! By breaking it down into manageable steps and understanding which tool to use, you'll master these problems quickly.
The Fundamental Difference: Why SUVAT Fails
The equations of motion (SUVAT) are only valid if acceleration \( a \) is constant. If the force \( F \) is variable, then by Newton’s Second Law (\( F = ma \)), the acceleration \( a \) must also be variable.
Recap: Newton's Second Law in Variable Form
The starting point for every variable force problem is still:
$$ F_{net} = ma $$
Since the force is variable, it will be expressed as a function of time \( t \), velocity \( v \), or displacement \( x \).
$$ F(t \text{ or } v \text{ or } x) = m a $$
The challenge then becomes choosing the correct mathematical definition for \( a \) that allows us to set up a separable differential equation (DE).
The Further Mechanics Toolbox: Three Forms of Acceleration
We rely on three calculus definitions of acceleration, and selecting the correct one is the key to setting up the problem successfully.
\( v = \frac{dx}{dt} \)
\( a = \frac{dv}{dt} \)
\( a = \frac{d^2x}{dt^2} \)
In Further Mechanics, we introduce a powerful chain rule identity derived from \( a = \frac{dv}{dt} \):
$$ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx} $$
Here are the three forms of acceleration you will use, based on what the variable force \( F \) depends on:
-
If \( F \) depends on Time (\( t \)): Use \( a = \frac{dv}{dt} \)
You want to find how velocity changes over time. Your setup will look like:
$$ F(t) = m \frac{dv}{dt} $$ -
If \( F \) depends on Velocity (\( v \)): Use \( a = \frac{dv}{dt} \)
You want to find how velocity changes over time. Your setup will look like:
$$ F(v) = m \frac{dv}{dt} $$ (Note: If you need to find displacement \( x \), you often solve for \( v(t) \) first, then use \( v = \frac{dx}{dt} \).) -
If \( F \) depends on Displacement (\( x \)): Use \( a = v \frac{dv}{dx} \)
You want to relate velocity directly to position without needing time \( t \). Your setup will look like:
$$ F(x) = m v \frac{dv}{dx} $$
Tool Selection Summary (Crucial!)
| If \( F \) is a function of... | Use this form for \( a \) | Resulting Differential Equation |
|---|---|---|
| \( t \) (Time) | \( \frac{dv}{dt} \) | \( \int dv = \int \frac{F(t)}{m} dt \) |
| \( v \) (Velocity) | \( \frac{dv}{dt} \) | \( \int \frac{m}{F(v)} dv = \int dt \) |
| \( x \) (Displacement) | \( v \frac{dv}{dx} \) | \( \int m v dv = \int F(x) dx \) |
Key Takeaway: Choosing the correct definition of \( a \) based on the variable defining \( F \) is the critical first step. It determines which variables you can separate and integrate.
The Solving Method: Separation of Variables
The syllabus restricts problems to differential equations where the variables are separable. This means you can rearrange the equation so all terms involving one variable (e.g., \( v \) and \( dv \)) are on one side, and all terms involving the other variable (e.g., \( t \) and \( dt \), or \( x \) and \( dx \)) are on the other.
Step-by-Step Procedure
- Identify all forces: Draw a diagram if needed and determine the net force \( F_{net} \).
- Formulate \( F = ma \): Write \( F_{net} \) as a function of \( t, v, \) or \( x \).
- Choose \( a \): Select the appropriate form of acceleration (\( \frac{dv}{dt} \) or \( v \frac{dv}{dx} \)).
-
Separate Variables: Algebraically manipulate the equation so that the differential terms (like \( dv \) or \( dx \)) are multiplied only by their corresponding variable functions.
Example Separation: If you have \( m \frac{dv}{dt} = 5v^2 \), you separate it to \( \frac{m}{5v^2} dv = dt \).
- Integrate: Integrate both sides. Don't forget the constant of integration (\( +C \)) on one side!
- Apply Initial/Boundary Conditions: Use the given starting conditions (e.g., at \( t=0, v=u \)) to find the value of the constant \( C \).
- Solve and Interpret: Use the resulting equation to solve for velocity, displacement, or time as required by the question.
Case Study 1: Forces Dependent on Velocity, \( F(v) \)
This is the most common scenario, often involving drag or resistance forces, which oppose motion and increase with speed.
Analogy: Skydiving
When a skydiver falls, the force of gravity is constant (\( mg \)), but air resistance \( R \) increases as their velocity \( v \) increases (often \( R \propto v \) or \( R \propto v^2 \)). The net downward force is \( F_{net} = mg - R(v) \).
Finding Velocity \( v \) as a function of Time \( t \)
Since \( F \) depends on \( v \), we use \( a = \frac{dv}{dt} \):
$$ F(v) = m \frac{dv}{dt} $$
If the resistance is proportional to velocity, \( R = kv \), and the particle is moving downwards:
$$ mg - kv = m \frac{dv}{dt} $$
Separating the variables:
$$ \frac{m}{mg - kv} dv = dt $$
Integrating both sides (this often involves integration of the form \( \int \frac{f'(x)}{f(x)} dx = \ln|f(x)| + C \)):
$$ \int \frac{m}{mg - kv} dv = \int dt $$
$$ -\frac{m}{k} \ln|mg - kv| = t + C $$
Did you know? Terminal Velocity
When the object reaches its terminal velocity (\( V \)), the acceleration is zero (\( a=0 \)), meaning the net force is zero. In the case above, this happens when \( F_{net} = mg - kV = 0 \), so \( V = \frac{mg}{k} \). You can often solve for terminal velocity without integration!
Finding Displacement \( x \) as a function of Velocity \( v \)
Sometimes, we want to find the distance travelled to reach a certain velocity. Here, we must be careful. We still start with \( F(v) = ma \), but substitute \( a = v \frac{dv}{dx} \).
$$ F(v) = m v \frac{dv}{dx} $$
Separating variables (note that \( v \) moves to the left side with \( dv \)):
$$ \frac{m v}{F(v)} dv = dx $$
$$ \int \frac{m v}{F(v)} dv = \int dx = x $$
Common Mistake to Avoid: When \( F \) is a function of \( v \), students sometimes incorrectly jump to using \( a = v \frac{dv}{dx} \) to find \( x \) directly. While mathematically possible, the integration can be much harder. It is often easier to find \( v(t) \) first, then integrate \( v = \frac{dx}{dt} \) to find \( x(t) \). Always check what the question is asking for!
Key Takeaway (Case 1): Velocity-dependent forces typically use \( a = \frac{dv}{dt} \) to find time, and require a second integration step (or the alternative form \( a = v \frac{dv}{dx} \)) to find displacement.
Case Study 2: Forces Dependent on Displacement, \( F(x) \)
This type of force usually relates to gravitational attraction that changes with distance, or simplified models of forces like those from Hooke's law (which often leads to simple harmonic motion, a concept explored elsewhere in mechanics).
Analogy: The Earth and Spacecraft
The force of gravity exerted by Earth on a satellite decreases as the distance \( x \) from the centre of the Earth increases (e.g., \( F \propto \frac{1}{x^2} \)).
Finding Velocity \( v \) as a function of Displacement \( x \)
Since \( F \) depends on \( x \), we instantly use \( a = v \frac{dv}{dx} \):
$$ F(x) = m v \frac{dv}{dx} $$
Separating the variables (this separation is very neat!):
$$ F(x) dx = m v dv $$
Integrating both sides:
$$ \int F(x) dx = \int m v dv $$
The right side always integrates nicely to \( \frac{1}{2} m v^2 \), which is related to kinetic energy. The integral on the left side calculates the work done by the variable force over the displacement \( x \).
This relationship demonstrates the Work-Energy Principle: Work Done = Change in Kinetic Energy.
Example Integration: If \( F(x) = 3x^2 \):
$$ \int 3x^2 dx = \int m v dv $$
$$ x^3 + C_1 = \frac{1}{2} m v^2 $$
Finding Time \( t \) as a function of Displacement \( x \)
If the question asks for the time taken to travel a distance \( x \), you must first find \( v(x) \) using the method above, and then substitute \( v = \frac{dx}{dt} \).
From the velocity equation, rearrange to get \( v \) in terms of \( x \):
$$ v = \sqrt{f(x)} $$
Substitute \( v = \frac{dx}{dt} \):
$$ \frac{dx}{dt} = \sqrt{f(x)} $$
Separate variables again:
$$ \frac{1}{\sqrt{f(x)}} dx = dt $$
Integrate:
$$ \int \frac{1}{\sqrt{f(x)}} dx = \int dt = t + C $$
Warning: This final integration step can be complex, often resulting in trigonometric or logarithmic functions, and sometimes inverse hyperbolic functions (though remember that calculus requirements are restricted to 9709 Pure Mathematics 3 content).
Key Takeaway (Case 2): Displacement-dependent forces almost always use \( a = v \frac{dv}{dx} \) first to find velocity as a function of position, leveraging the work-energy connection.
Quick Review and Problem-Solving Tricks
Encouraging Note:
Remember, the complexity here is not in the mechanics itself (it’s just \( F=ma \)), but in the resulting differential equation. If you are strong in integration and separation of variables from Pure Mathematics 3, you are well-equipped!
Choosing the Right Starting Point
Always ask yourself: "What variable is the net force a function of?"
- If \( F = f(t) \) or \( F = f(v) \), start with \( F = m \frac{dv}{dt} \).
- If \( F = f(x) \), start with \( F = m v \frac{dv}{dx} \).
Common Errors to Avoid
- Forgetting the Constant of Integration: Always include \( +C \) after your first integration. Use the initial conditions (e.g., \( t=0, v=u \)) immediately to find \( C \).
- Incorrect Separation: Ensure \( dx \) is only multiplied by functions of \( x \), \( dv \) only by functions of \( v \), and \( dt \) only by functions of \( t \).
- Missing the Mass \( m \): In \( F = ma \), students sometimes forget to divide \( F \) by \( m \) before integrating the acceleration variable (e.g., leaving \( \int F(v) dv \) instead of \( \int \frac{m}{F(v)} dv \)).
- Using \( a = v \frac{dv}{dx} \) when Time is Required: If the question explicitly asks for the time taken, you must involve \( t \) at some point, usually via \( a = \frac{dv}{dt} \), or by substituting \( v = \frac{dx}{dt} \) into your final velocity expression.
Summary: Linear Motion Under Variable Force
This chapter links Newton’s Second Law directly with the power of calculus to analyze realistic motion where forces are not constant. The whole process relies on formulating the correct separable differential equation using the definitions \( a = \frac{dv}{dt} \) or \( a = v \frac{dv}{dx} \). Mastering the selection of the correct form is 80% of the battle! Good luck!